Class 9 · ChemistrySet 2 · Answer key

Answer Key – Set 2 ഉത്തരസൂചിക

Step-by-step answers and the marking scheme for every question of Set 2, for teachers and parents to value the paper. Accept any correct answer in the student’s own words; give method marks even when the final number is wrong.

ANSWER KEY & MARKING SCHEME · SET 2

Standard IX · Chemistry · Annual Examination

Total score: 60 · Time: 2 hours

Section AAnswer all questions · 10 × 1 = 10 marks

The scientist who proposed that electrons revolve round the nucleus only in certain definite orbits, each with a definite energy, is

(a) J. J. Thomson(b) Ernest Rutherford(c) Niels Bohr(d) James Chadwick

[1]

Step-by-step solution

In the Bohr atom model (ബോർ ആറ്റം മാതൃക, 1913) electrons revolve in definite orbits, and each orbit has a definite energy – the energy levels (ഊർജനിലകൾ).
Thomson proposed the plum pudding model, Rutherford the solar system (nuclear) model, and Chadwick discovered the neutron.

✅ Answer: (c) Niels Bohr

Marking: 1 mark for (c).

📖 Revise: Ch 1 · Bohr model · Chapter 1 – Structure of the Atom

An atom has the electron configuration 2, 5 and 7 neutrons in its nucleus. Its atomic number is and its mass number is .

[1]

Step-by-step solution

Number of electrons = 2 + 5 = 7. In a neutral atom, protons = electrons, so the atomic number Z = 7 (nitrogen).
Mass number A = protons + neutrons = 7 + 7 = 14.

✅ Answer: Atomic number 7; mass number 14

Marking: ½ mark for each blank.

📖 Revise: Ch 1 · Atomic number and mass number · Chapter 1 – Structure of the Atom

Helium, neon and argon normally do not take part in chemical reactions because

(a) they are gases at room temperature(b) the outermost shell of their atoms is stable (8 electrons; 2 in helium)(c) their atoms have no electrons in the outermost shell(d) all of them are in period 1

[1]

Step-by-step solution

Noble gases (ഉൽകൃഷ്ട വാതകങ്ങൾ) have 8 electrons in the outermost shell (helium has 2, which fills its only shell).
This is a stable arrangement, so they normally do not take part in chemical reactions.

✅ Answer: (b) the outermost shell of their atoms is stable (8 electrons; 2 in helium)

Marking: 1 mark for (b).

📖 Revise: Ch 2 · Noble gases · Chapter 2 – Periodic Table

Pick the odd one out and give the reason: scandium (2, 8, 9, 2), titanium (2, 8, 10, 2), calcium (2, 8, 8, 2), vanadium (2, 8, 11, 2).

[1]

Step-by-step solution

Calcium has 2 outermost electrons and 8 in the shell inside it – it is in group 2, an alkaline earth metal (main group).
In scandium, titanium and vanadium the new electrons go into the shell just inside the outermost shell (9, 10, 11) – they are transition elements (സംക്രമണ മൂലകങ്ങൾ).

✅ Answer: Calcium – it is a main group element (group 2, alkaline earth metal); the others are transition elements.

Marking: ½ mark for calcium, ½ mark for the reason.

📖 Revise: Ch 2 · Transition elements · Chapter 2 – Periodic Table

Aluminium burns in oxygen: 4Al + 3O₂ → 2Al₂O₃. Which statement is correct?

(a) Aluminium atoms gain electrons.(b) Oxygen is oxidised.(c) Aluminium is the reducing agent.(d) Oxygen is the reducing agent.

[1]

Step-by-step solution

Each aluminium atom loses 3 electrons: Al → Al³⁺ + 3e⁻. Losing electrons is oxidation, so aluminium is oxidised.
Each oxygen atom gains 2 electrons: O + 2e⁻ → O²⁻, so oxygen is reduced.
Aluminium gives the electrons that reduce oxygen, so aluminium is the reducing agent (നിരോക്സീകാരി); oxygen is the oxidising agent.

✅ Answer: (c) Aluminium is the reducing agent.

Marking: 1 mark for (c).

📖 Revise: Ch 4 · Oxidising and reducing agents · Chapter 4 – Redox Reactions

When reactants are heated, the reaction usually becomes faster. The main reason is that

(a) the threshold energy of the reaction becomes lower(b) more reactant particles get a kinetic energy equal to or greater than the threshold energy(c) the number of reactant particles in the vessel increases(d) the reactant particles become larger

[1]

Step-by-step solution

Heating increases the kinetic energy of the particles.
So more particles reach the threshold energy (ത്രെഷോൾഡ് എനർജി) and there are more effective collisions every second – the rate increases.
Heating does not lower the threshold energy, and it does not change the number or size of the particles.

✅ Answer: (b) more reactant particles get a kinetic energy equal to or greater than the threshold energy

Marking: 1 mark for (b).

📖 Revise: Ch 5 · Temperature and threshold energy · Chapter 5 – Chemical Kinetics

Find the relation and fill in: Haber process : iron :: contact process :

[1]

Step-by-step solution

In the Haber process, N₂ + 3H₂ ⇌ 2NH₃, iron is the catalyst.
In the contact process, 2SO₂ + O₂ ⇌ 2SO₃, the catalyst (ഉൽപ്രേരകം) is vanadium pentoxide, V₂O₅.

✅ Answer: vanadium pentoxide (V₂O₅)

Marking: 1 mark (name or formula).

📖 Revise: Ch 5 · Catalysts in industry · Chapter 5 – Chemical Kinetics

Hollow, cage-like spheres made of carbon atoms arranged in pentagons and hexagons are called

(a) graphene(b) buckyballs (fullerene)(c) graphite(d) lamp black

[1]

Step-by-step solution

Fullerene (ഫുള്ളറീൻ) is a crystalline allotrope of carbon. Its hollow spheres of pentagons and hexagons are called buckyballs; cylinders are buckytubes.
Graphene is a single flat layer of graphite; lamp black is an amorphous form.

✅ Answer: (b) buckyballs (fullerene)

Marking: 1 mark for (b).

📖 Revise: Ch 7 · Allotropes of carbon · Chapter 7 – Non-metals

Which of these is a use of carbon monoxide, not of carbon dioxide?

(a) fire extinguishers(b) making soda water(c) reducing agent in metallurgy(d) dry ice for stage “smoke”

[1]

Step-by-step solution

Carbon dioxide is used in fire extinguishers, soda water and soft drinks, and as dry ice for stage "smoke".
Carbon monoxide is used as a gaseous fuel (water gas, producer gas) and as a reducing agent in metallurgy.

✅ Answer: (c) reducing agent in metallurgy

Marking: 1 mark for (c).

📖 Revise: Ch 7 · Carbon dioxide and carbon monoxide · Chapter 7 – Non-metals

10.

The IUPAC name of CH≡C–CH₂–CH₂–CH₂–CH₃ is

(a) hexane(b) hexene(c) hexyne(d) pentyne

[1]

Step-by-step solution

Count the carbon atoms: 6 → word root hex.
There is a triple bond (C≡C), so the suffix is yne: hex + yne = hexyne. (Hexane has only single bonds; hexene has a double bond.)

✅ Answer: (c) hexyne

Marking: 1 mark for (c).

📖 Revise: Ch 8 · IUPAC naming · Chapter 8 – Organic Chemistry

Section BAnswer all questions · 8 × 2 = 16 marks

11.

An atom of potassium is written as 3919K. (a) Find the number of protons, electrons and neutrons in it. (b) Write its electron configuration. How many electrons are there in its outermost shell, and which shell (K, L, M or N) is it?

[2]

Step-by-step solution

Given: mass number A = 39, atomic number Z = 19
Formula: protons = electrons = Z; neutrons = A − Z
Substitution: neutrons = 39 − 19
Answer: (a) 19 protons, 19 electrons, 20 neutrons
(b) Fill the shells: K 2, L 8, M 8 (for elements up to 20 the outermost shell takes at most 8), and the last electron goes to N. Electron configuration (ഇലക്ട്രോൺ വിന്യാസം) = 2, 8, 8, 1 → 1 electron in the outermost N shell.

✅ Answer: (a) 19 protons, 19 electrons, 20 neutrons (b) 2, 8, 8, 1 – one electron in the outermost N shell

Marking: (a) protons and electrons ½, neutrons ½; (b) configuration ½, one electron in the N shell ½.

📖 Revise: Ch 1 · A − Z · Ch 1 · Electron configuration · Chapter 1 – Structure of the Atom

12.

The electron configurations of four elements are given (the letters are not their real symbols): P 2, 8, 8 · Q 2, 8, 8, 2 · R 2, 7 · S 2, 8, 1. Find the group number of each and name the family to which it belongs.

[2]

Step-by-step solution

Groups 1 and 2: group number = number of outermost electrons. Groups 13–18: group number = outermost electrons + 10.
P: 8 outermost electrons → 8 + 10 = group 18 – noble gases (ഉൽകൃഷ്ട വാതകങ്ങൾ).
Q: 2 outermost electrons → group 2 – alkaline earth metals (ആൽക്കലൈൻ എർത്ത് ലോഹങ്ങൾ).
R: 7 outermost electrons → 7 + 10 = group 17 – halogens (ഹാലൊജനുകൾ).
S: 1 outermost electron → group 1 – alkali metals (ആൽക്കലി ലോഹങ്ങൾ).

✅ Answer: P – group 18, noble gases; Q – group 2, alkaline earth metals; R – group 17, halogens; S – group 1, alkali metals

Marking: ½ mark for each element (group and family both correct).

📖 Revise: Ch 2 · Finding the group · Ch 2 · Families · Chapter 2 – Periodic Table

13.

Use the electronegativity values of Fig 3.12 (Si 1.90, Cl 3.16, K 0.82, Br 2.96). Find the electronegativity difference in (a) silicon tetrachloride, SiCl₄ (b) potassium bromide, KBr. Say whether each compound is ionic or covalent.

[2]

Step-by-step solution

Given: Si 1.90, Cl 3.16, K 0.82, Br 2.96
Formula: difference = larger value − smaller value; 1.7 or more → generally ionic, less than 1.7 → covalent
Substitution: (a) 3.16 − 1.90 (b) 2.96 − 0.82
Answer: (a) 1.26, less than 1.7 → SiCl₄ is covalent. (b) 2.14, more than 1.7 → KBr is ionic.

✅ Answer: (a) 1.26 – covalent (b) 2.14 – ionic

Marking: Each part: difference ½, ionic/covalent ½.

📖 Revise: Ch 3 · Electronegativity and the 1.7 rule · Chapter 3 – Chemical Bonding

14.

Say whether each change is oxidation or reduction. Give the reason.
(a) Mg → Mg²⁺ + 2e⁻ (b) Br + e⁻ → Br⁻ (c) Cu²⁺ + 2e⁻ → Cu (d) S²⁻ → S + 2e⁻

[2]

Step-by-step solution

Oxidation (ഓക്സീകരണം) is the loss of electrons; reduction (നിരോക്സീകരണം) is the gain of electrons.
(a) The magnesium atom loses 2 electrons → oxidation.
(b) The bromine atom gains 1 electron → reduction.
(c) The copper ion gains 2 electrons and becomes a copper atom → reduction.
(d) The sulphide ion loses 2 electrons and becomes a sulphur atom → oxidation.

✅ Answer: (a) oxidation (b) reduction (c) reduction (d) oxidation

Marking: ½ mark for each (answer with the loss/gain reason).

📖 Revise: Ch 4 · Oxidation and reduction · Chapter 4 – Redox Reactions

15.

(a) Sort these into fast and slow reactions: inflating of a car air bag; rusting of an iron gate in the monsoon; white precipitate forming when silver nitrate solution is added to salt solution; weathering of rocks. (b) Give one reason why it is good for us that rusting of iron is a slow reaction.

[2]

Step-by-step solution

Reactions take place at very different rates (രാസപ്രവർത്തനവേഗം).
(a) Fast: the air bag fills with nitrogen in about 0.03 s; the white precipitate of silver chloride forms at once.
Slow: rusting of iron takes days to months; weathering of rocks takes many years.
(b) If rusting were very fast, iron bridges, vehicles, gates and tools would be destroyed in a very short time.

✅ Answer: (a) Fast: air bag, silver chloride precipitate. Slow: rusting, weathering. (b) If rusting were fast, iron bridges, vehicles and tools would be destroyed quickly.

Marking: (a) all four correct 1 mark (two or three correct ½); (b) 1 mark.

📖 Revise: Ch 5 · Rate of reaction · Chapter 5 – Chemical Kinetics

16.

Ammu adds sugar to a glass of lime juice and stirs. After a while, however much she stirs, some sugar stays at the bottom; the temperature does not change. (a) What kind of solution is the lime juice now? Why? (b) How can she make the remaining sugar dissolve without adding any more water or juice? (c) She does this, and then lets the glass cool slowly without disturbing it. No sugar separates. What kind of solution is it now?

[2]

Step-by-step solution

(a) Saturated solution (പൂരിത ലായനി) – the maximum amount of sugar that can dissolve at this temperature has dissolved, so no more dissolves.
(b) Warm the juice. The solubility of a solid like sugar generally increases with temperature, so the extra sugar dissolves.
(c) It now holds more sugar than is needed to saturate it at room temperature – a supersaturated solution (അതിപൂരിത ലായനി). It is unstable: a small crystal or a shake will make the extra sugar separate.

✅ Answer: (a) saturated (b) warm it (c) supersaturated

Marking: (a) saturated with reason 1; (b) ½; (c) ½.

📖 Revise: Ch 6 · Saturated, unsaturated and supersaturated solutions · Chapter 6 – Solutions

17.

Ravi has a white powder and wants to know whether it is a carbonate. He dissolves it in water and adds barium chloride solution: a white precipitate forms. (a) What does this show? (b) What must he do next to confirm that the salt is a carbonate, and what will he see?

[2]

Step-by-step solution

(a) A carbonate solution gives a white precipitate of barium carbonate with barium chloride, so the salt may be a carbonate.
(b) Add dilute hydrochloric acid to the precipitate. If it dissolves with effervescence (bubbles of carbon dioxide, which turn lime water milky), the salt is a carbonate.

✅ Answer: (a) The salt may be a carbonate (white barium carbonate forms). (b) Add dilute HCl to the precipitate – it dissolves with effervescence of CO₂ → carbonate.

Marking: (a) 1 mark; (b) adding dilute HCl ½, dissolving with effervescence ½.

📖 Revise: Ch 7 · Test for carbonates (Table 7.3) · Chapter 7 – Non-metals

18.

(a) An alkene has 9 carbon atoms. How many hydrogen atoms does it have? Write its molecular formula and IUPAC name. (b) Do the same for an alkyne with 8 carbon atoms.

[2]

Step-by-step solution

Given: alkene: n = 9; alkyne: n = 8
Formula: alkenes C_nH_2n; alkynes C_nH_2n−2
Substitution: (a) 2 × 9 = 18 (b) 2 × 8 − 2 = 14
Answer: (a) 18 hydrogen atoms – C₉H₁₈, nonene (non + ene). (b) 14 hydrogen atoms – C₈H₁₄, octyne (oct + yne).

✅ Answer: (a) 18 H – C₉H₁₈, nonene (b) 14 H – C₈H₁₄, octyne

Marking: Each part: number of H ½, formula and name ½.

📖 Revise: Ch 8 · General formulae · Ch 8 · Word roots and suffixes · Chapter 8 – Organic Chemistry

Section CAnswer all questions · 6 × 3 = 18 marks

19.

The figure shows an incomplete orbit diagram of a silicon atom; only the electrons of the K shell are drawn. (a) Write the atomic number and mass number of silicon. (b) Copy the diagram and complete it by drawing the electrons in the L and M shells. Write the electron configuration. (c) According to Bohr, which of the shells K, L and M has the highest energy? If an electron moves from the L shell to the M shell, is energy absorbed or given out?

[3]

Step-by-step solution

(a) Atomic number = number of protons = 14. Mass number = protons + neutrons = 14 + 14 = 28.
(b) 14 electrons: K holds 2, L holds 8, and the remaining 14 − 2 − 8 = 4 go to M. Configuration 2, 8, 4.
Completed diagram:
(c) The energy of a shell increases with its distance from the nucleus, so M has the highest energy. Moving to a higher shell (L → M), the electron absorbs energy (ഊർജം ആഗിരണം ചെയ്യുന്നു).

✅ Answer: (a) Z = 14, A = 28 (b) 2, 8, 4 (diagram) (c) M has the highest energy; energy is absorbed

Marking: (a) ½ + ½; (b) 8 electrons in L and 4 in M drawn ½, configuration ½; (c) M ½, absorbed ½.

📖 Revise: Ch 1 · Orbit diagrams · Ch 1 · Bohr model – energy levels · Chapter 1 – Structure of the Atom

20.

Meera’s teacher shows four compounds: nickel sulphate (green), potassium dichromate (orange), cobalt nitrate (pink-red) and common salt, sodium chloride (white). (a) Name the element in each coloured compound that belongs to the transition elements. (b) Why is sodium chloride not coloured? (c) The elements of groups 3 to 12 are called transition elements. Why?

[3]

Step-by-step solution

(a) Most compounds of transition elements are coloured (നിറമുള്ള സംയുക്തങ്ങൾ): nickel sulphate – nickel; potassium dichromate – chromium; cobalt nitrate – cobalt.
(b) Sodium (group 1) and chlorine (group 17) are main group elements. Sodium chloride has no transition element, so it is white.
(c) Groups 3–12 lie between the highly metallic elements of group 2 and the less metallic elements from group 13 onwards. They show this gradual change (transition) in properties, so they are called transition elements.

✅ Answer: (a) nickel, chromium, cobalt (b) it contains no transition element (c) they show a gradual change (transition) from the highly metallic group 2 to the less metallic group 13 onwards

Marking: (a) all three 1 mark (two correct ½); (b) 1 mark; (c) 1 mark.

📖 Revise: Ch 2 · Transition elements and coloured compounds · Chapter 2 – Periodic Table

21.

Ammonia (NH₃) is a covalent compound (N: Z = 7; H: Z = 1). (a) Write the electron configuration of nitrogen. How many more electrons does it need to complete its octet? (b) Draw the electron dot diagram of an NH₃ molecule (use dots for the electrons of nitrogen and crosses for those of hydrogen). (c) How many single bonds are there in NH₃? How many electrons are there around the nitrogen atom after bonding?

[3]

Step-by-step solution

(a) Nitrogen: 2, 5. It has 5 outermost electrons, so it needs 3 more to complete the octet.
(b) Nitrogen shares one electron pair with each of three hydrogen atoms – each pair has one electron from N (dot) and one from H (cross). The two remaining electrons of N stay as a lone pair. Electron dot diagram (ഇലക്ട്രോൺ ഡോട്ട് ഡയഗ്രം):
(c) Three shared pairs → 3 single bonds (ഏകബന്ധനം). Around nitrogen: 3 shared pairs (6) + 1 lone pair (2) = 8 electrons – an octet. Each hydrogen has 2 electrons, like helium.

✅ Answer: (a) 2, 5 – needs 3 more (b) dot diagram (c) 3 single bonds; 8 electrons around N

Marking: (a) configuration ½, 3 electrons ½; (b) correct diagram 1; (c) 3 bonds ½, 8 electrons ½.

📖 Revise: Ch 3 · Covalent compounds – dot diagrams · Ch 3 · Single bonds · Chapter 3 – Chemical Bonding

22.

Boxes P and Q have the same volume and are at the same temperature. They contain the particles of two gases that react with each other. (a) Which box has the higher concentration of reactants? What is meant by concentration here? (b) In which box will the reaction be faster? Explain using collision theory. (c) Not every collision between the particles gives the products. Which collisions do? What is threshold energy?

[3]

Step-by-step solution

(a) Q. Concentration (ഗാഢത) is the number of reactant particles (molecules) per unit volume; Q has 14 particles and P only 6 in the same volume.
(b) Q. By collision theory (കൊളിഷൻ സിദ്ധാന്തം) a reaction takes place when reactant particles collide. With more particles in the same space, there are more collisions every second and so more effective collisions – the rate is higher.
(c) Only effective collisions (ഫലവത്തായ കൂട്ടിമുട്ടലുകൾ) – collisions between particles that have enough energy – give products. The threshold energy is the minimum kinetic energy that the reactant particles must have to change into products.

✅ Answer: (a) Q (b) Q – more particles per unit volume → more collisions → more effective collisions (c) effective collisions, with at least the threshold energy

Marking: (a) Q with meaning 1; (b) Q ½, explanation ½; (c) effective collisions ½, threshold energy ½.

📖 Revise: Ch 5 · Collision theory · Ch 5 · Concentration · Chapter 5 – Chemical Kinetics

23.

In an experiment at 20 °C, 9 g of common salt was just enough to saturate 25 g of water; no more would dissolve. (a) Find the solubility of common salt at 20 °C. (b) Find the mass percentage of salt in this saturated solution. (c) What is the least mass of water at 20 °C needed to dissolve 18 g of common salt completely?

[3]

Step-by-step solution

Given: 9 g of salt saturates 25 g of water at 20 °C
Formula: solubility (ലേയത്വം) = mass of solute that saturates 100 g of solvent; mass % = mass of solute ÷ mass of solution × 100
Substitution: (a) 9 ÷ 25 × 100 (b) 9 ÷ (9 + 25) × 100 = 9 ÷ 34 × 100 (c) 18 ÷ 36 × 100
Answer: (a) solubility = 36 g per 100 g of water (b) mass percentage ≈ 26.5 % (c) 50 g of water
(c) Reason: 36 g of salt needs 100 g of water, so 18 g (half of 36 g) needs half of 100 g = 50 g.

✅ Answer: (a) 36 g (per 100 g of water) (b) about 26.5 % (c) 50 g

Marking: (a) 1 mark; (b) mass of solution ½, answer ½; (c) 1 mark.

📖 Revise: Ch 6 · Solubility · Ch 6 · Mass percentage · Chapter 6 – Solutions

24.

The figure shows the laboratory preparation of carbon dioxide. Dilute hydrochloric acid is poured through A on to the solid B. (a) Name the apparatus A and the substance B. (b) Write the balanced equation for the reaction in the flask. (c) Why is the gas collected in an upright jar C? What will you see when the gas is passed into clear lime water?

[3]

Step-by-step solution

(a) A is a thistle funnel (തിസിൽ ഫണൽ); B is marble chips (calcium carbonate, CaCO₃).
(b) CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂↑ (Ca 1 = 1, C 1 = 1, O 3 = 3, H 2 = 2, Cl 2 = 2 – balanced).
(c) Carbon dioxide is denser than air, so it sinks to the bottom of the upright jar and pushes the air up and out. When it is passed into clear lime water (ചുണ്ണാമ്പുവെള്ളം), the lime water turns milky.

✅ Answer: (a) A – thistle funnel; B – marble (calcium carbonate) (b) CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂ (c) CO₂ is denser than air; lime water turns milky

Marking: (a) ½ + ½; (b) correct formulae ½, balanced ½; (c) denser than air ½, turns milky ½.

📖 Revise: Ch 7 · Preparation of carbon dioxide · Chapter 7 – Non-metals

Section DAnswer all questions · 4 × 4 = 16 marks

25.

In the science club, Anu compares water (H₂O) and carbon dioxide (CO₂). Electronegativity values (Fig 3.12): H 2.20, C 2.55, O 3.44. (a) Find the electronegativity difference of the O–H bond and of the C=O bond. Are these bonds ionic or covalent? (b) In the O–H bond, which atom gets the partial negative charge (δ−)? Draw a water molecule and mark δ+ and δ−. (c) Both molecules have polar bonds, yet water is a polar molecule and carbon dioxide is non-polar. Why? (d) Water has a low molecular mass, but it is a liquid at room temperature. Which attraction between water molecules is responsible?

[4]

Step-by-step solution

Given: H 2.20, C 2.55, O 3.44
Formula: difference = larger value − smaller value; less than 1.7 → covalent
Substitution: O–H: 3.44 − 2.20 C=O: 3.44 − 2.55
Answer: (a) O–H 1.24, C=O 0.89. Both are less than 1.7, so both bonds are covalent (polar covalent).
(b) Oxygen is more electronegative, so it pulls the shared pair towards itself and gets δ−; each hydrogen gets δ+:
(c) In O=C=O the two oxygen atoms pull equally in opposite directions, so the pulls cancel – because of its shape the CO₂ molecule is non-polar. In the water molecule the pulls do not cancel, so water is polar.
(d) Hydrogen bonding (ഹൈഡ്രജൻ ബന്ധനം): the δ+ hydrogen of one water molecule is attracted by the δ− oxygen of another. These attractions hold the molecules together, so water is a liquid.

✅ Answer: (a) O–H 1.24, C=O 0.89 – both covalent (b) oxygen is δ− (c) shape of CO₂: the two pulls cancel (d) hydrogen bonding

Marking: (a) each difference ½ (nature included); (b) oxygen δ− ½, labelled diagram ½; (c) 1 mark; (d) 1 mark.

📖 Revise: Ch 3 · Polar molecules, shape, hydrogen bonding · Chapter 3 – Chemical Bonding

26.

Bordeaux mixture is sprayed on coconut palms and arecanut trees. Rajan dissolved copper sulphate in water in an old iron bucket and left it overnight. Next morning the blue colour had become paler and a reddish-brown coating of copper had formed on the inside of the bucket: Fe + CuSO₄ → FeSO₄ + Cu. (a) Write the equations showing the change in an iron atom and in a copper ion (Cu²⁺). (b) Which is oxidised and which is reduced? Give the reason. (c) Name the oxidising agent and the reducing agent. (d) Why did the blue colour become paler? Why do farmers prepare copper sulphate solution in plastic vessels, not in iron ones?

[4]

Step-by-step solution

(a) Iron atom: Fe → Fe²⁺ + 2e⁻. Copper ion: Cu²⁺ + 2e⁻ → Cu. (The sulphate ion, SO₄²⁻, does not change.)
(b) Iron loses electrons → iron is oxidised (ഓക്സീകരിക്കപ്പെടുന്നു). The copper ion gains electrons → it is reduced. (Oxidation numbers: Fe 0 → +2, increase; Cu +2 → 0, decrease.)
(c) Oxidising agent (ഓക്സീകാരി): Cu²⁺ (copper sulphate) – it takes electrons from iron and is itself reduced. Reducing agent (നിരോക്സീകാരി): iron – it gives electrons and is itself oxidised.
(d) The blue copper sulphate is used up: its copper ions change into copper metal (the brown coating), and pale-green iron sulphate forms, so the blue fades. In an iron vessel the iron itself reacts – the bucket is eaten away and the copper sulphate needed for spraying is wasted. Plastic does not react.

✅ Answer: (a) Fe → Fe²⁺ + 2e⁻; Cu²⁺ + 2e⁻ → Cu (b) iron oxidised, copper ion reduced (c) oxidising agent Cu²⁺ (copper sulphate); reducing agent Fe (d) copper sulphate is used up; iron reacts and the bucket is damaged

Marking: (a) ½ + ½; (b) oxidised ½, reduced ½ (with the electron reason); (c) ½ + ½; (d) ½ + ½.

📖 Revise: Ch 4 · Oxidising and reducing agents · Ch 4 · Loss and gain of electrons · Chapter 4 – Redox Reactions

27.

For the school science fair, Nithya studies the solubility of three salts in water. The table and the graph give the solubility in grams per 100 g of water.

Temperature (°C)	10	20	40	60	80
KNO₃	21	32	62	106	167
NH₄Cl	24	37	41	55	66
NaCl	36	36	36	37	37

(a) Which salt is more soluble at 20 °C: potassium nitrate or ammonium chloride? Which is more soluble at 60 °C? (b) What is the lowest temperature at which 62 g of potassium nitrate can dissolve completely in 100 g of water? (c) 55 g of ammonium chloride is dissolved in 100 g of water at 60 °C. Is the solution saturated or unsaturated? If it is heated to 80 °C, how much more ammonium chloride can it dissolve? (d) Which salt’s solubility changes the least with temperature? If a saturated solution of it in 100 g of water is cooled from 80 °C to 20 °C, about how much salt separates?

[4]

Step-by-step solution

(a) At 20 °C: KNO₃ 32 g, NH₄Cl 37 g → ammonium chloride is more soluble. At 60 °C: KNO₃ 106 g, NH₄Cl 55 g → potassium nitrate. (The curves cross between 20 °C and 40 °C.)
(b) The solubility of KNO₃ reaches 62 g at 40 °C. Below 40 °C less than 62 g dissolves in 100 g of water.
(c) At 60 °C the solubility of NH₄Cl is 55 g, so 55 g in 100 g of water is a saturated solution (പൂരിത ലായനി). At 80 °C the solubility is 66 g, so it can dissolve 66 − 55 = 11 g more.
(d) Sodium chloride – its curve is almost flat (36–37 g). Saturated at 80 °C: 37 g; at 20 °C only 36 g can stay dissolved, so 37 − 36 = about 1 g separates as crystals (പരലീകരണം).

✅ Answer: (a) 20 °C: ammonium chloride (37 g > 32 g); 60 °C: potassium nitrate (106 g > 55 g) (b) 40 °C (c) saturated; 11 g more (d) sodium chloride; about 1 g

Marking: (a) ½ + ½; (b) 1 mark; (c) saturated ½, 11 g ½; (d) NaCl ½, 1 g ½.

📖 Revise: Ch 6 · Solubility curves · Ch 6 · Saturated solutions · Chapter 6 – Solutions

28.

In a chemistry class in Kochi, the teacher draws two hydrocarbons, P and Q. (a) Write the molecular formula of P and of Q. What do you notice? (b) Write the IUPAC names of P and Q. (c) Q belongs to a homologous series. Name the series and write its general formula. Draw the structural formula of the member that comes just before Q. (d) Is P an alicyclic or an aromatic compound? Give the reason.

[4]

Step-by-step solution

(a) P: 5 carbons; hydrogens 1 + 1 + 2 + 2 + 2 = 8 → C₅H₈. Q: 5 carbons; hydrogens 1 + 0 + 2 + 2 + 3 = 8 → C₅H₈. Both have the same molecular formula, but one is a ring and the other an open chain.
(b) P: a ring (വലയസംയുക്തങ്ങൾ) of 5 carbons with one double bond → cyclo + pentene = cyclopentene. Q: 5 carbons (pent) with a triple bond (yne) → pentyne.
(c) Q is an alkyne (ആൽക്കൈനുകൾ); general formula (പൊതുവാക്യം) C_nH_2n−2. In a homologous series (ഹോമലോഗസ് ശ്രേണി) neighbours differ by –CH₂–, so the member just before C₅H₈ is C₄H₆, butyne:
(d) Alicyclic (ആലിസൈക്ലിക് സംയുക്തങ്ങൾ): P is a ring compound that is like an open-chain alkene closed into a ring. It is not a benzene-type ring with a characteristic smell (aromatic).

✅ Answer: (a) both C₅H₈ (b) P cyclopentene, Q pentyne (c) alkynes, C_nH_2n−2; butyne C₄H₆ (d) alicyclic

Marking: (a) both formulae ½, observation ½; (b) ½ + ½; (c) series and general formula ½, butyne structure ½; (d) alicyclic with reason 1.

📖 Revise: Ch 8 · Ring compounds · Ch 8 · Homologous series · Chapter 8 – Organic Chemistry

Question numbers by chapter

Chapter	A (1)	B (2)	C (3)	D (4)	Total
1. Structure of the Atom	1, 2	11	19	–	7
2. Periodic Table	3, 4	12	20	–	7
3. Chemical Bonding	–	13	21	25	9
4. Redox Reactions	5	14	–	26	7
5. Chemical Kinetics	6, 7	15	22	–	7
6. Solutions	–	16	23	27	9
7. Non-metals	8, 9	17	24	–	7
8. Organic Chemistry	10	18	–	28	7
Total	10	16	18	16	60

Score to grade

Kerala grading scale: A+ 54–60 (90 %+), A 48–53, B+ 42–47, B 36–41, C+ 30–35, C 24–29, D+ 18–23, D 12–17, E below 12. For every question the student lost marks on, the linked guide page explains the idea again.